README.md

1695. Maximum Erasure Value

You are given an array of positive integers nums and want to erase a subarray containing unique elements. The score you get by erasing the subarray is equal to the sum of its elements.

Return the maximum score you can get by erasing exactly one subarray.

An array b is called to be a subarray of a if it forms a contiguous subsequence of a, that is, if it is equal to a[l],a[l+1],...,a[r] for some (l,r).

Example 1:

Input: nums = [4,2,4,5,6]
Output: 17
Explanation: The optimal subarray here is [2,4,5,6].

Example 2:

Input: nums = [5,2,1,2,5,2,1,2,5]
Output: 8
Explanation: The optimal subarray here is [5,2,1] or [1,2,5].

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 104

Solutions (Ruby)

1. Two Pointers

# @param {Integer[]} nums
# @return {Integer}
def maximum_unique_subarray(nums)
  i = 0
  counter = {}
  counter.default = 0
  sum = 0
  ret = 0

  (0...nums.size).each do |j|
    counter[nums[j]] += 1
    sum += nums[j]
    while counter[nums[j]] > 1
      counter[nums[i]] -= 1
      sum -= nums[i]
      i += 1
    end
    ret = [ret, sum].max
  end

  ret
end

Solutions (Rust)

1. Two Pointers

use std::collections::HashMap;

impl Solution {
    pub fn maximum_unique_subarray(nums: Vec<i32>) -> i32 {
        let mut i = 0;
        let mut counter = HashMap::new();
        let mut sum = 0;
        let mut ret = 0;

        for j in 0..nums.len() {
            *counter.entry(nums[j]).or_insert(0) += 1;
            sum += nums[j];
            while *counter.get(&nums[j]).unwrap() > 1 {
                *counter.get_mut(&nums[i]).unwrap() -= 1;
                sum -= nums[i];
                i += 1;
            }
            ret = ret.max(sum);
        }

        ret
    }
}